log graphs


Ask a Question, Get an Answer!
Hundreds of tutors are online and ready to help you right now!
Graph of Logarithmic Function
Graph of Logarithmic Function - For x (0, ), the value of log x is uniquely defined.For x (0, ), the value of log x is uniquely defined. \ x g log x is a well-defined function from (0, ) to (- , ). The value of e to one place of decimal i..
Graph of Logarithmic Function
For x (0, ), the value of log x is uniquely defined.For x (0, ), the value of log x is uniquely defined. \ x g log x is a well-defined function from (0, ) to (- , ). The value of e to one place of decimal i..
Logarithmic function
Logarithmic function is f (x) = log x. Its graph i..
Draw the graph of y = log 4x.
Draw the graph of y = log 4 x ...
Which of the following transformations has to be applied to the graph ..
Which of the following transformations has to be applied to the graph of f ( y ) = log y to get the graph of g ( y ) = log 9 y ? => Vertical shrink, by the factor 1 log 9 or Vertical stretch, by the factor log 9 or Vertical stretch, by the..
Find the y-intercept of the graph of y = log4(x + 4).
Find the y -intercept of the graph of y = log 4 ( x + 4). => 0 or 1 or 3 or 2..
Which of the following transformations has to be applied to the graph ..
Which of the following transformations has to be applied to the graph of f ( x ) = log x to get the graph of g ( x ) = log 1 2 x ? => Reflection of the graph of f ( x ) about x - axis followed by a horizontal shrink by a fac..
Identify the equation of the graph shown.
Identify the equation of the graph shown. => y = - log x + 9 or y = - 9log x or y = log x - 9 or y = log x + 9..
Identify the equation of the graph shown.
Identify the equation of the graph shown. => y = log x or y = - log x or y = 2log x or y = - 2log x..
Find the x-intercept of the graph of y = 4(x - 1) - 2.
Find the x -intercept of the graph of y = 4 ( x - 1) - 2. => log 4 2 + 1 or log 4 2 - 1 or log 2 4 - 1 or log 2 4 + 1..
Result Pages   :     1     2     3


See what our Users say :
Very fast and clear. Made sure I understood the concepts instead of giving the answers to the problem.
This Tutor Vista is GREAT! loved this session, it helped me heaps.
This was excellent. I was cluesless on this subject and now I completely understand it.Mind blowing service,Thank you Tutor Vista
I really like this tutoring, and this is much easier now that I understand what I'm doing. Thanks again -Tina

Looking for More Help!

Popular Help Topics
Math Help     Math Homework Help     Math Word Problems      Chemistry Homework Help    Trigonometry Formulas     Precalculus Help
Algebra 1     Solving Square Root     Algebra Word Problems   Science Homework Help       Simplifying Fractions        Trigonometry Help
Pre Algebra  Math Answers               Math Problems                 Algebra Homework Help       Math Questions                 Homework Help
Algebra Help  Calculus Help              Statistics Help                  Chemistry Help                     Algebra 2 Help